Answer to Question #175577 in Linear Algebra for william

Question

solve the system of three variable linear equations.

x = 3z - 5

2x + 2z = y + 16

7x - 5z = 3y + 19

Answer 1

x = 3z-5 ....(1)

2x+2z = y+16 ....(2)

7x-5z=3y+19 ....(3)

Multiplying through equation (2) by -3, we have

-6x-6z= -3y-48 ....(4)

Adding equation (3) and (4), we have

x-11z = -29

x = 11z-29 .... (5)

Equating equation (1) and (5), we have

3z-5 = 11z-29

3z-11z= -29+5

-8z = -24

z = 3

Substituting the value of z into equation (1), we have

x = 3(3)-5 = 9-5 = 4

Substituting the value of x and z into (2), we have

2(4)+2(3)=y+16

8+6=y+16

14 =y+16

y= -2

So therefore, x=4, y=-2,z=3